JEE Advanced 2019 · Paper 1 · Chemistry Q5

ChemistryMultiple correct+4 / -1Question 23 of 54

A tin chloride Q undergoes the following reactions (not balanced)

$\mathbf Q + \ce{Cl-} \to \mathbf X$

$\mathbf Q + \ce{Me3N} \to \mathbf Y$

$\mathbf Q + \ce{CuCl2} \to \mathbf Z + \ce{CuCl}$

X is a monoanion having pyramidal geometry. Both Y and Z are neutral compounds. Choose the correct option(s)

The central atom in X is

sp^3

hybridized

There is a coordinate bond in Y

The oxidation state of the central atom in Z is +2

The central atom in Z has one lone pair of electrons

Answer and worked solution

Official answer: A, B

Identify Q. Q is $\ce{SnCl2}$ : a Lewis acid with a lone pair on Sn, and a reducing agent.

X is $\ce{SnCl3-}$ : three bond pairs and one lone pair, so it is pyramidal and Sn is $sp^3$ . (A) is true.
Y is $\ce{SnCl2.NMe3}$ : the N lone pair is donated to Sn, a coordinate bond. (B) is true.
Z. $\ce{SnCl2 + 2CuCl2 -> SnCl4 + 2CuCl}$ , so Z is $\ce{SnCl4}$ . Sn is +4 and has no lone pair, so (C) and (D) are false.

(A), (B)

Topics:Group 14 VSEPR theory Hybridisation

JEE Advanced 2019 · Paper 1, Chemistry question 5